Question

Consider the following statements
I. The modulus function $f:R\to R$ given by $f(x)=|x|$ is neither one-one nor onto.
II. The signum function $f:R\to R$ given by $f(x)=\{\begin{array}{ll}1,& \text{ if }x>0\\ 0,& \text{ if }x=0\\ -1,& \text{ if }x<0\end{array}$ is bijective.
III. The function $f:N\to N$ defined by $f(n)=\{\begin{array}{ll}\frac{n+1}{2},& \text{ if }n\text{ is odd }\\ \frac{n}{2},& \text{ if }n\text{ is even }\end{array}$ for all $n\in N$ is not bijective.
Choose the correct option.

• A. I and II are true
• B. II and III are true
• C. I and III are true
• D. All are true

Answer 1

Answer: (C)

I. Here, $f:R\to R$ is given by $f(x)=|x|=\{\begin{array}{ll}x,& \text{ if }x\ge 0\\ -x,& \text{ if }x<0\end{array}$
It is seen that $f(-1)=|-1|=1,f(1)=|1|=1$
Therefore, $f(-1)=f(1)$ but $-1\ne 1$
Therefore, $f$ is not one-one.
Now, consider $-1\in R$
It is known that $f(x)=|x|$ is always non-negative.
Thus, there does not exist any element $x$ in domain $R$ such that
$f(x)=|x|=-1.$
Therefore, $f$ is not onto.
Hence, the modulus function is neither one-one nor Onto.
II. $f:R\to R,f(x)=\{\begin{array}{llc}1,& \text{ if }& x>0\\ 0,& \text{ if }& x=0\\ -1,& \text{ if }& x<0\end{array}$
It is seen that $f(1)=f(2)=1$ but $1\ne 2$. Therefore, $f$ is not one-one.
Now, as $f(x)$ takes only three values (1,0 or $-1$ ), therefore for the element $-2$ in codomain $R$, there does not exist any $x$ in domain $R$ such that $f(x)=-2$.
Therefore, $f$ is not onto.
Hence, the Signum function is neither one-one nor onto.
III. Here, $f:N\to N$ is defined as $f(n)=\{\begin{array}{ll}\frac{n+1}{2},& \text{ if }n\text{ is odd }\\ \frac{n}{2},& \text{ if }n\text{ is even }\end{array}$ for all $n\in N$. It can be observed that $f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1$ (by definition of $f$ ) $\therefore f(1)=f(2)$, where $1\ne 2$.
Therefore, $f$ is not one-one. Consider a natural number $n$ in codomain $N$.
Case I When $n$ is odd.
Therefore, $n=2r+1$ for some $r\in W$.
Then, there exists $4r+1\in N$ such that
$f(4r+1)=\frac{4r+1+1}{2}=2r+1$
Therefore, $f$ is onto.
Case II When $n$ is even
Therefore, $n=2r$ for some $r\in N$
Then, there exists $4r\in N$ such that $f(4r)=\frac{4r}{2}=2r$
Therefore, $f$ is onto. Hence, $f$ is not a bijective function.